Problem: Let $h(x)=\dfrac{4x^2}{x^2-1}$ Where does $h$ have critical points? Choose all answers that apply: Choose all answers that apply: (Choice A) A $x=-1$ (Choice B) B $x=0$ (Choice C) C $x=1$ (Choice D) D $h$ has no critical points.
A critical point of $h$ is a point in the domain of $h$ where the derivative is either equal to zero or undefined. So in order to find the critical points of $h$, let's find its derivative. $\begin{aligned} h'(x)&=\dfrac{d}{dx}\left[ \dfrac{4x^2}{x^2-1} \right] \\\\ &=\dfrac{(x^2-1)\cdot \dfrac{d}{dx}[4x^2]-4x^2\cdot \dfrac{d}{dx}[x^2-1]}{(x^2-1)^2} \\\\ &=\dfrac{(x^2-1)\cdot 8x-4x^2\cdot 2x}{(x^2-1)^2} \\\\ &=\dfrac{-8x}{(x^2-1)^2} \end{aligned}$ Now let's look for $x$ -values where $h'$ is zero or undefined. $\dfrac{-8x}{(x^2-1)^2}=0$ at $x=0$. $\dfrac{-8x}{(x^2-1)^2}$ is undefined at $x=-1$ and $x=1$. However, $h(x)=\dfrac{4x^2}{x^2-1}$ is also undefined at $x=1$ and $x=-1$ so they aren't critical points. In conclusion, this is the only $x$ -value where $h$ has a critical point: $x=0$